What is the general solution for -19sin(θ)+19/2 =0 ?

The general solution for -19sin(θ)+19/2 =0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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-19sin(θ)+19/2 =0

- 19 sin (θ) + \frac{19}{2} = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

- 19 sin (θ) + \frac{19}{2} = 0

Move

\frac{19}{2}

to the right side

- 19 sin (θ) = - \frac{19}{2}

Divide both sides by

- 19

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

y, e^{8 x} = sin (x + 3 y)

π cos (π x) + π sin (π x) = 0

144 = 100 + 6.9 3^{2} - 2 \cdot 10 \cdot 6.93 \cdot cos (x)

0.09 = 0.18 sin (1.082 t)

cos (2 x) = sec (x)

What is the general solution for -19sin(θ)+19/2 =0 ?
The general solution for -19sin(θ)+19/2 =0 is θ= pi/6+2pin,θ=(5pi)/6+2pin